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In the Dür, 2000 paper, he gave a statement that

(...) Any state $\vert\psi\rangle$ can be obtained from Bell State with certainty

From his paper too, it's known that we can transform $\vert\psi\rangle$ and $\vert\phi\rangle$ into each other with certainty by means of LOCC iff they are related by local unitaries (LU). So, I can write the transformation down such that (correct me if I'm wrong)

$U_1 \otimes U_2 \vert\Phi^+\rangle = \alpha_{00}\vert00\rangle + \alpha_{01}\vert01\rangle + \alpha_{10}\vert10\rangle + \alpha_{11}\vert11\rangle \\ \left(U_1 \otimes U_2\right)\left(\frac{1}{\sqrt{2}}\left(\vert00\rangle + \vert11\rangle\right)\right) = \alpha_{00}\vert00\rangle + \alpha_{01}\vert01\rangle + \alpha_{10}\vert10\rangle + \alpha_{11}\vert11\rangle$

From this step, I find it difficult how to obtain the LU transformation.

Michele Amoretti
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Steve J.
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1 Answers1

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What you want is the theory of the Schmidt decomposition. Let's say you have a two-qubit state $|\psi\rangle$ which is written $$ |\psi\rangle=\alpha_{00}|00\rangle+\alpha_{01}|01\rangle+\alpha_{10}|10\rangle+\alpha_{11}|11\rangle, $$ then it's helpful to construct a matrix $$ C=\left(\begin{array}{cc} \alpha_{00} & \alpha_{01} \\ \alpha_{10} & \alpha_{11} \end{array}\right). $$ This matrix $C$ has a singular value decomposition, meaning there exist unitaries $U$ and $V$ such that $$ C=UDV^\dagger, $$ where $D$ is diagonal with positive semi-definite entries.

You will find that $$ (U^\dagger \otimes V^\dagger )|\psi\rangle=|\phi\rangle $$ where $|\phi\rangle$ is the Bell state. Hence, really what you want is to know how to do the singular value decomposition. If you're not familiar with it, it can help to relate it to the more famliar eigenvalues: $$ CC^\dagger=UD^2U^\dagger,\qquad C^\dagger C=VD^2V^\dagger. $$

Note that $|\phi\rangle$, the Bell state, already has $U=V=I$ when written in this way.

DaftWullie
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